Cartesian Coordinates Plane Strain
16 If the deformation of a cylindrical body is such that there is no axial components of the displacement and that the other components do not depend on the axial coordinate, then the body is said to be in a state of plane strain. If e3 is the direction corresponding to the cylindrical axis, then we have
and the strain components corresponding to those displacements are
Ox2
1 (dui du2
and the nonzero stress components are Tii, Ti2, T22, T33 where
17 Considering a static stress field with no body forces, the equilibrium equations reduce to:
dTii + dTi2
dxi dx2 dTi2 + 8X22 dxi dx2 dTss dxi
we note that since T33 = T33(x\, x2), the last equation is always satisfied.
is Hence, it can be easily verified that for any arbitrary scalar variable $, if we compute the stress components from
d2$ 
(10.17)  
T11= 
dx2  
d2$ 
(10.18)  
T22 = 
dx2  
d 2$ 
(10.19)  
T12 = 
dxidx2 
then the first two equations of equilibrium are automatically satisfied. This function $ is called Airy stress function.
19 However, if stress components determined this way are statically admissible (i.e. they satisfy equilibrium), they are not necessarily kinematically admissible (i.e. satisfy compatibility equations).
20 To ensure compatibility of the strain components, we obtain the strains components in terms of $ from Hooke's law, Eq. 5.1 and Eq. 10.15.
Eii E22 Ei2
dx2 j
21 For plane strain problems, the only compatibility equation, 4.140, that is not automatically satisfied is d2Eu d2E22 n d2 E12
dxidx2
thus we obtain the following equation governing the scalar function $
or d4$ d4 $ d4$ T7 + 2jn^2 + TT =0 or V4$ = 0 dxi dxfdx2 dx4
Hence, any function which satisfies the preceding equation will satisfy both equilibrium and kinematic and is thus an acceptable elasticity solution.
22 We can also obtain from the Hooke's law, the compatibility equation 10.21, and the equilibrium equations the following
23 Any polynomial of degree three or less in x and y satisfies the biharmonic equation (Eq. 10.23). A systematic way of selecting coefficients begins with
24 The stresses will be given by
oo oo
J2J2n(n  1)CmnXmyn2
25 Substituting into Eq. 10.23 and regrouping we obtain oo oo
J2 J2[(m+2)(m+1)m(m1)Cm+2,u2+2m(m1)n(n1)Cmn+(n+2)(n+1)n(n1)Cm2,u+2]xm2yn2 = 0
but since the equation must be identically satisfied for all x and y, the term in bracket must be equal to zero.
(m+2)(m +1)rn(rn1)Cm+2,n2 +2m(m1)n(n1)Cmn + (n+2)(n +1)n(n1)Cm_2,n+2 = 0 (10.28)
Hence, the recursion relation establishes relationships among groups of three alternate coefficients which can be selected from
0 
0 
C02 
C03 
C04 
C05 C06 • • •  
0 
C11 
C12 
C13 
C14 
C15 • • •  
C20 
C21 
C22 
C23 
C24  
C30 
C31 
C32 
C33 
(10.29)  
C40 
C41 
C42  
C50 
C51  
C60 
(4)(3)(2)(1)C4o + (2)(2)(1)(2)(1)C22 + (4)(3)(2)(1)Co4 = 0
10.2.1.1 Example: Cantilever Beam
26 We consider the homogeneous fourthdegree polynomial
$4 = C40X4 + C31x3y + C22x2y2 + C\3xy3 + C04 y4 with 3C40 + C22 + 3Co4 = 0,
27 The stresses are obtained from Eq. 10.26a10.26c
Txx = 2C22X2 +6Ci3xy + 12Co4 y2 Tyy = I2C40X2 +6C3ixy + 2C22 y2
These can be used for the endloaded cantilever beam with width b along the z axis, depth 2a and length L.
28 If all coefficients except C13 are taken to be zero, then
29 This will give a parabolic shear traction on the loaded end (correct), but also a uniform shear traction Txy = —3C13a2 on top and bottom. These can be removed by superposing uniform shear stress Txy = +3C13a2 corresponding to $2 = —3C13a2xy. Thus
note that C20 = C02 = 0, and C11 = —3C13a2. 30 The constant C13 is determined by requiring that hence and the solution is
T Tx
4a3b
3P P
bxy 3P
4abxy 4a3b xy
T xy
2a3b
31 We observe that the second moment of area for the rectangular cross section is I = 5(2a)3/12 = 2a3b/3, hence this solution agrees with the elementary beam theory solution
Cllxy H Cl3xy
4abxy 4a3b xy
10.2.2 Polar Coordinates 1G.2.2.1 Plane Strain Formulation
32 In polar coordinates, the strain components in plane strain are, Eq. 9.46
and the equations of equilibrium are
33 Again, it can be easily verified that the equations of equilibrium are identically satisfied if
+ r2 dO2 
(10.41)  
Tee = 
d2$ dr2 
(10.42)  
Tre = 
d dr 
( 1 d$ \ \r~30) 
(10.43) 
34 In order to satisfy the compatibility conditions, the cartesian stress components must also satisfy Eq. 10.24. To derive the equivalent expression in cylindrical coordinates, we note that Tn + T22 is the first scalar invariant of the stress tensor, therefore
r dr r2 dO2 or2
35 We also note that in cylindrical coordinates, the Laplacian operator takes the following form
dr2 r dr r2 BO2
Thus, the function $ must satisfy the biharmonic equation
10.2.2.2 Axially Symmetric Case
37 If $ is a function of r only, we have
r dr and
dr4 r dr3 r2 dr2 r3 dr 38 The general solution to this problem; using Mathematical
DSolve[phi''''[r]+2 phi'''[r]/rphi''[r]/r"2+phi'[r]/r"3==0,phi[r],r]
39 The corresponding stress field is
and the strain components are (from Sect. 9.8.1)
duT dr
+ (1  3v  4v2)B + 2(1  v  2v2)Blnr + 2(1  v  2v2)C
+ (3  v  4v2)B + 2(1  v  2v2)B lnr + 2(1  v  2v2)C
4r0B
(1 + v)A  (1 + v)Br + 2(1  v  2v2)r ln rB + 2(1  v  2v2)rC
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